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Ksju [112]
3 years ago
7

72 yo m with sob for 3 days. hx of htn and cad. receiving 2 l/min of o2 via nasal cannula. temp 37 c, p 110/min, resp 20/min, bp

150/80. bilateral crackles and wheezes. hct is 28%, leukocyte count is 8000. pulmonary artery cath shows cardiac index of 2 l/min (n = 2.5 – 4.2) and a pulmonary artery occlusion pressure of 28. abg shows ph 7.49; pco2 30; po2 58. most appropriate next step in management
Chemistry
1 answer:
Ksivusya [100]3 years ago
8 0
<span>I believe the answer is B)Diuretic therapy
</span>
The patient has alkalosis(pH 7.49), increased respiratory rate (20/min) and decreased carbon dioxide level(<span>pCO2 30). These findings point to respiratory alkalosis. 
The reason for increased respiratory rate probably because of heart failure that is shown by the decreased </span>cardiac index, pO2, <span>crackles, and wheezes. The treatment for congestive heart failure is to decrease the preload and blood pressure by giving a strong diuretic like furosemide.</span>
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The reaction between sodium and chlorine that forms table salt is shown
Olegator [25]

Answer: 2 Na (s) + Cl(g) -> 2 NaCl (s)

Explanation:

7 0
3 years ago
A certain substance X has a normal freezing point of -6.4 C and a molal freezing point depression constant Kf= 3.96 degrees C.kg
Brut [27]

Answer:  1.0\times 10^2g

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=T_f^0-T_f=(-6.4-(13.6))^0C=7.2^0C = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte like urea)

K_f = freezing point constant = 3.96^0C/m

m= molality

\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}}\times \text{weight of solvent in kg}}

Weight of solvent (X)= 950 g = 0.95 kg  

Molar mass of non electrolyte (urea) = 60.06 g/mol

Mass of non electrolyte (urea) added = ?

7.2=1\times 3.96\times \frac{xg}{60.06 g/mol\times 0.95kg}

x=1.0\times 10^2g

Thus 1.0\times 10^2g urea was dissolved.

8 0
3 years ago
I begin the reaction with 0.45 g of beryllium. If my actual yield of beryllium chloride (mm = 79.91 g/mol) was 3.5 grams, what w
trasher [3.6K]

The percentage of yield was 777.78%

<u>Explanation:</u>

We have the equation,

Be [s]  +  2 HCl [aq]  →  BeCl 2(aq]  + H 2(g]  ↑ Be (s]  + 2 HCl [aq]  →  BeCl 2(aq]  + H 2(g] ↑

To find the percent yield we have the formula

Percentage of Yield= what you actually get/ what you should theoretically get  x 100

                                   =3.5 g/0.45 g 100

                                    = 777.78 %

The percentage of yield was 777.78%

5 0
3 years ago
Escribe aquí algo que consideres que requiere un balance para que se lleve a cabo correctamente:
Galina-37 [17]

Answer:

Direct weighing means that an object is placed directly on a balance and the mass read. Weighing directly requires that the balance be carefully zeroed (reads zero with nothing on the balance pan) in order to obtain accurate results.

Explanation:

Hope I Help?

PLease Brainly Me!

3 0
3 years ago
A tank of gas is found to exert 8.6 atm at 38°C. What would be the required
Vesna [10]

Answer:

36.2 K

Explanation:

Step 1: Given data

  • Initial pressure of the gas (P₁): 8.6 atm
  • Initial temperature of the gas (T₁): 38°C
  • Final pressure of the gas (P₂): 1.0 atm (standard pressure)
  • Final temperature of the gas (T₂): ?

Step 2: Convert T₁ to Kelvin

We will use the following expression.

K = °C +273.15

K = 38 °C +273.15 = 311 K

Step 3: Calculate T₂

We will use Gay Lussac's law.

P₁/T₁ = P₂/T₂

T₂ = P₂ × T₁/P₁

T₂ = 1.0 atm × 311 K/8.6 atm = 36.2 K

6 0
2 years ago
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